Osmosis investigation. My prediction is that as the concentration of the solute increases, net flow will increase out the potato

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Osmotic potential in potatoes.

Introduction:

Osmosis is the diffusion of water from a high concentration to a low concentration through a semi-permeable membrane. This process does not require energy (passive transport) as it works through pressure to reach equilibrium, i.e. the same concentration on both sides, and in this case the solute concentration would be equal and the solutions would be isotonic at the end- in this case there is no net flow of water through the membrane.

There are two main types of solutions in regards to osmosis.

e.g's

-A hypotonic solution is a solution with a lower concentration of solutes compared to another. A cell in a hypotonic solution has a higher concentration of solutes than the surrounding solution, causing water to flow into the cell. Cells can burst (cytolysis) in some cases.

-A hypertonic solution is a solution with a higher concentration of solutes compared to another. A cell in a hypertonic solution has a lower concentration of solutes than its surroundings, leading to a decrease of water in the cell. Water will leave the cell by means of osmosis, this can lead to plasmolysis, whereby a cell loses turgor pressure and the protoplasm (yellow in diagram) peels away from the cell wall. The effect of this is a plant will wilt if too many cells lose turgor.

The factors affecting osmosis rate are:

Temperature.

The molecules in the solutions move faster with a higher temperature, therefore the frequency of collisions increases meaning quicker osmosis.

Concentration gradients between solutes (osmotic gradient)

[Basically] Because a hypotonic solution has a smaller solute concentration, there is a higher water concentration, which means there are more collisions with the membrane than the hypertonic solution causing a faster net flow to the hypertonic solution.

Surface area to volume

  • A small cell has a large SA:V ratio and therefore can exchange molecules with it's external solution rapidly, because there are more H2O collisions with the membrane due to the high ratio, however the probability of successful osmosis is not increased. Ions and water can be lost by diffusion and osmosis, and heat can be lost at a faster rate.
  • A large cell has a small SA:V ratio and therefore exchanges molecules with it's external solution slowly, because there are less collisions with the membrane due to the smaller ratio.

The factor I chose:

I choose osmotic gradient as my independent variable, because it's the easiest variable from which to produce reliable, accurate and reproducible results. I didn't chose SA:V as the independent variable, because of the time it takes to prepare that amount of potato to meet a good standard precision, it would simply take to long to repeat the experiment for reliability. I could have chosen to use temperature as a variable, however I felt it would be wiser to chose otherwise, because if I used a tube in a water bath it would take too long to perform the experiment, as it would take too long for the tube to heat up and if I simply put hot water in a tube and waited for it to cool down, that too would take too long. Additionally the membrane may be altered by the heat, which technically is unfair to the test, because the membrane should be a controlled variable. So, I chose concentration.

Hypothesis and Prediction:

My prediction is that as the concentration of the solute increases, net flow will increase out the potato. At first I hope to see that the net flow will be into the potato until equilibrium is reached, after this point as concentration increases the net flow will be out the potato. I believe this because of particle theory- that the solution with the highest concentration of water molecules will diffuse through the partially permeable membrane faster than the other side due to the greater amount of water molecules colliding with the membrane; this applies to my prediction- as the lowest concentration of water is in the potato at first the water will flow in and as concentration of water in the solution decreases the two solutions will reach equilibrium. After that, the outer solution will have a smaller concentration of water thus the potato will lose water, thus losing water weight (similarly the potato will gain water weight at the beginning too).

Method

Preliminary Work:

Plan:

Equipment:

  • Test tube rack
  • 5 Test tubes
  • 3 Measuring cylinders
  • 1 bottle of distilled water (bottles to be shared, amount to use: 50ml)
  • 1 bottle of sucrose solution (bottles to be shared, amount to use: 50ml)
  • 1 or more potatoes, enough for 5 pieces
  • 1 core-borer (size to be used unknown as it depends on which one I can aquire)
  • 1 scalpel
  • 1 cutting tile
  • Paper towel
  • 1 stop watch
  • Weighing scales

Techniques:

First I will set up the test tubes in the rack and then I will prepare the solutions.

1. I chose 3 measuring cylinders for this purpose; I will use two of them to quantify the sucrose solution and the distilled water, leaving the third to add the two mixtures together. This extra effort saves time and keeps precision, as there is no need to make repeat solutions as every thing is correct before they are mixed leaving an accurate and precise mixture. To make my solutions of 0, 0.25, 0.5, 0.75 and 1 mole sucrose, I will use ratios- e.g. for 0.5:  1:1, so 10ml of each.

Next I will prepare the potato pieces for their "soak".

2. I will aim to attain all the pieces from one potato. I will take the potato lay it on the tile and bore downwards with the tool. Once I have 5 separate cylinders of potato I will roll them on the paper towel to soak the excess moisture. Thereafter I will weigh them and cut them down till they are 1.00 gram each. Part of this technique is cutting of slivers from the cross section of the cylinder so that they maintain a similar surface area. The next step will be to put the pieces in each solution, while my partner starts the watch on the time the first piece is added.

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While the potatoes soak, I will replace the equipment.

3. Once the 20 min time period for the potatoes is up I will drain all the tubes in a manner so that the potato pieces are still in their tubes. I will take them to the scales and one at a time I will roll them on paper towel to soak up excess (undiffused) moisture and then weigh them.

Once I have the results I will write them into my pre-drawn table, then dispose of the potato.

The results of the preliminary experiment are as follows:

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